(a) Suppose that $N=n$. What is the conditional distribution (name and parameter(s)) of
$T$?
(a) Exponential distribution with rate $pλ$
Under the condition that $N=n$, no sock until the $n$-th cause the system to fail, and so $T$ measures the time until the $n$-th shock; ie, its distribution is that for the sum of $n$ exponential random variables each with rate $\lambda$.
(b) Suppose that a failed system is always immediately replaced by a new system. What is the distribution (name and parameter(s)) of the number of replacements that occur during a fixed time interval $[0,t]$?
(b) Poisson distribution with parameter $λt$
That is the distribution for the count of shocks within the interval.
We seek the distribution for the count of failure causing shocks within the interval.
(c) Suppose that $5$ shocks occur during $[0,t]$. What is the distribution (name and parameter(s)) of the number of replacements during $[0,t]$?
(c) Binomial distribution ~ $(5,p)$
$\color{green}\checkmark$ That is the distribution for the count of failure causing shocks among $5$ shocks which occur independently with identical rate $p$.
(d) Given that $T=t$, what is the conditional distribution (name and parameter(s)) of $N$?
(d) Poisson distribution with parameter $λt$
At time $T$ the first failure-causing shock has occurred, and $N-1$ counts the shocks which have not caused failure.
What is the distribution for the count of non-failure-causing shocks within period $[0,t)$ ?
Your definition of expected value is at least non-standard, and may be wrong. For a continuous random variable $X$ with probability density function $f$,
$$E(X)=\int_{-\infty}^\infty xf(x)\ dx$$
Given the CDF, the PDF can be easily found by
$$f(x) = \frac{d}{dx}F(x)$$
So in your specific example, we have
$$f(t) = \lambda e^{-\lambda t}, t>0$$
$$E(T) = \int_0^\infty t\lambda e^{-\lambda t}\ dt=\frac{1}{\lambda}$$
Now, we can use Bayes' rule to find a CDF for the random variable with conditional.
$$F_{T\geq\tau}(t)=P(T\leq t|T\geq \tau) = \frac{P(\tau \leq T\leq t)}{P(T\geq \tau)} = \frac{F(t)-F(\tau)}{1-F(\tau)},t\geq\tau$$
$$=\frac{e^{-\lambda \tau}-e^{-\lambda t}}{e^{-\lambda \tau}}$$
And we can differentiate to get a new PDF
$$f_{T\geq\tau}(t) = \frac{\lambda e^{-\lambda t}}{e^{-\lambda \tau}}=\lambda e^{-\lambda (t-\tau)},t>\tau$$
And finally the expected value is
$$E(T) = \int_\tau^\infty t\lambda e^{-\lambda (t-\tau)}\ dt=\int_0^\infty (u+\tau)\lambda e^{-\lambda (u)}\ du=\tau+\frac{1}{\lambda}$$
This gives a very nice result for the exponential distribution shown, owing to its memoryless property. However this process should work (albeit with more difficulty) for other distributions.
Best Answer
Yes, exactly. The computation of $\int_t^{t+\delta} p(t)\, dt = F(t+\delta)-F(t)$ is the probability that a brand new component is going to fail during the interval $[t,t+\delta]$. And this decreases with larger $t$ precisely because the component is less likely to survive up to time $t$ in the first place.
Bayes' Theorem says $$P(A \mid B) = \frac{P(B \mid A) P(A)}{P(B)}$$
If $A$ is the event $t \leq T \leq t+\delta$ and $B$ is the event $t \leq T$, then $P(B\mid A) = 1$: if $A$ is true then $B$ is necessarily true. We know $P(A) = F(t+\delta)-F(t)$ as explained in your previous question. And $P(B) = 1 - F(t)$ since $F(t)$ is the probability $T \leq t$, $P(B)$ is the probability $T \geq t$, and the probability the real number $T$ is exactly any particular $t$ is zero. So
$$ P(t \leq T \leq t + \delta \mid t \leq T) = \frac{F(t+\delta)-F(t)}{1-F(t)} = \frac{e^{-\lambda t} - e^{-\lambda (t+\delta)}}{e^{-\lambda t}} = 1 - e^{-\lambda \delta} $$
That's right, this probability doesn't depend on $t$ at all. The exponential distribution models a case where an old component is just as likely to fail in any given time interval as a brand new component. We see this a lot in particle physics: particles of the same type act just the same, and how long that particle has existed isn't relevant. A carbon-14 nucleus in a dinosaur fossil and a carbon-14 nucleus formed today in the upper atmosphere have the same chance of decaying in the next year.